我尝试编写以下函数
typedef struct node{
char *name;
int n; //number of kids
struct node **kids;
int dtype; // COMPOSITE or BASIC
union data{
double price; //for BASIC
char *quantity; //for COMPOSITE
}data;
}node;
void create_kid(node **parent){
if(*parent == NULL){
(*parent) = malloc(sizeof(node));
(*parent)->n = 0;
(*parent)->kids = NULL;
}else{
(*parent)->n += 1;
(*parent)->kids = realloc((*parent)->kids, ((*parent)->n)*(sizeof(node)));
(*parent)->kids[(*parent)->n - 1] = malloc(sizeof(node));
(*parent)->kids[(*parent)->n - 1]->n = 0;
(*parent)->kids[(*parent)->n - 1]->kids = NULL;
}
}
void insert_c(node *node, char *str){
node->name = malloc(MAX);
node->name = str;
node->dtype = COMPOSITE;
}
void insert_b(node *node, char *str){
node->name = malloc(MAX);
node->name = str;
node->dtype = BASIC;
}
从这样的输入中制作出这样的 n 叉树
BIKE(2*WHEEL(RIM[60.0 ],
2*AXLE,
SPOKE[120.],
HUB(2*GEAR[25.],AXLE(5*BOLT[0.1], 7 * NUT[.15]))),
FRAME(REARFRAME [175.00],
1*FRONTFRAME (FORK[22.5] ,AXLE, 2 *HANDLE[10.])))
我知道 n 元树的另一种表示形式,包括 *siblings 和 *firstkid,但我相信这种 **kids 表示形式更适合我的情况。我想问几件事。
首先,功能有什么问题吗?它们适合构建 n 叉树吗?
其次,即使我得到了正确的功能,我也无法从输入中构造树。例如,根据输入,如果我一次读一个单词,函数调用与我的函数必须是这样的:
create_kid(&tree);
insert_c(tree, "BIKE");
create_kid(&tree);
insert_c(tree->kids[0], "WHEEL");
create_kid(&(tree->kids[0]));
insert_c(tree->kids[0]->kids[0], "RIM");
.
.
.
create_kid(&tree);
insert_c(tree->kids[1], "FRAME");
.
.
如果要以这种方式形成树,我需要从较高的深度到较低的深度,从
NUT
至FRAME
例如。所以我相信一定有更简单的方法,也许是递归的方法。有什么办法可以通过递归来做到这一点?
最佳答案
先说小事:
如果 AXLE 节点的同一个实例位于“树”中的不同位置,则 n 叉树不再是 n 叉树。它是一个图表。如果您的车轴还包含一辆自行车,那么它甚至会是一个循环图。
在实现所有这些时,您会注意到不同之处。您将需要引用计数或图表中的某些内容来决定何时真正释放节点。在真正的 n 元树中,您不需要这样做,因为每个节点在树中只出现一次。
接下来的行政事项:
您决定使用指针数组作为数据结构来包含节点的子节点。因此,为了降低代码的复杂性,作为第一步,实际创建一个(动态)指针数组是一个好主意。好处:您的树代码没有动态数组代码的负担,您可以进行单元测试,甚至可以将您的动态指针数组重新用于其他项目。
我敲了一个示例指针数组实现(没有真正深入测试)来展示原理。
为了让这里显示的代码更“用户友好”——首先是我使用的包含,以便将其排除在外:
#include <crtdbg.h> // Windows specific - helps find memory leaks.
#include <stdint.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
现在到指针数组的实现:
typedef struct PointerArray_tag
{
void**data;
size_t capacity;
size_t size;
} PointerArray_t;
void PointerArrayInitialize(PointerArray_t *array)
{
array->data = NULL;
array->capacity = 0;
array->size = 0;
}
int PointerArrayReserve(PointerArray_t *array, size_t capacity)
{
if (array->capacity < capacity)
{
void **newData = (void**)realloc(array->data, capacity * sizeof(void*));
if (NULL != newData)
{
array->data = newData;
array->capacity = capacity;
return 1; // 1 = "true" in C and indicates success.
}
return 0; // 0 = "false" in C and indicates failure...
}
else
{
return 1;
}
}
typedef void (*PointerArrayElementFree_t)(void *element);
int PointerArrayResize(PointerArray_t *array, size_t newSize, void* value, PointerArrayElementFree_t elementFree)
{
if (NULL == array) return 0;
if (NULL == elementFree) return 0;
if (newSize < array->size)
{
for (size_t index = newSize; index < array->size; index++)
{
elementFree(array->data[index]);
}
array->size = newSize;
return 1;
}
else if (newSize > array->size)
{
size_t oldSize = array->size;
if (PointerArrayReserve(array, newSize))
{
for (size_t index = oldSize; index < newSize; index++)
{
array->data[index] = value;
}
array->size = newSize;
return 1;
}
return 0; // Reserve() failed, so this function also failed.
}
else
return 1; // oldSize == newSize - nothing to do.
}
int PointerArrayInitializeWithCapacity(PointerArray_t *array, size_t initialCapacity)
{
PointerArrayInitialize(array);
return PointerArrayReserve(array, initialCapacity);
}
int PointerArrayPushBack(PointerArray_t *array, void*element)
{
if (PointerArrayReserve(array, array->size + 1))
{
array->data[array->size] = element;
array->size++;
return 1;
}
return 0;
}
int PointerArrayClear(PointerArray_t *array, PointerArrayElementFree_t elementFree)
{
if (NULL == elementFree) return 0;
if (NULL == array) return 0;
for (size_t index = 0; index < array->size; index++)
{
elementFree(array->data[index]);
array->data[index] = NULL;
}
array->size = 0;
return 1;
}
int PointerArrayUninitialize(PointerArray_t *array, PointerArrayElementFree_t elementFree)
{
if (PointerArrayClear(array, elementFree))
{
free(array->data);
array->data = NULL;
array->capacity = 0;
return 1;
}
return 0;
}
这里没什么特别的。具有通常操作的 void 指针的动态数组。
接下来,虽然不是 100% 适合您的问题(我觉得这样做会花费我更多的时间),但 n-ary 树实现。如上所述,它还没有准备好用作图表。
另外,我冒昧地假设了一种语法形式:
<Material> ::= <Count> <Name> <Size>
| <Count> <Name>
解析这个产品,解析器首先会找到一个数字(例如 2 个轮子),然后是 Material 的名称(轮子),然后是可选的测量值(例如 Nuts[0.15])。
当应用程序使用 AST 并且过于复杂时,更接近您的问题的替代语法稍后将更难处理:
<Material> ::= <Count> <Name> <Size>
| <Count> <Name>
| <Name>
| <Name> <Size>
这暗示如果您对输入文本的语法有影响,您可以对其进行一些改进。
所以,这里是 ComponentTree 代码,它具有上述约束。
enum NodeContentType
{
EMPTY = 0
, NUMBER = 1
, NAME = 2
};
typedef struct NodeContent_tag
{
NodeContentType type;
union
{
float number;
char *name;
};
} NodeContent_t;
typedef struct ComponentNode_tag
{
NodeContent_t content;
ComponentNode_tag * parent;
PointerArray_t children;
} ComponentNode_t;
void ComponentNodeFree(ComponentNode_t *node)
{
// Clean up anything heap based in NodeContent_t.
switch (node->content.type)
{
case NAME:
free(node->content.name);
node->content.name = NULL;
node->content.type = EMPTY;
break;
default:
// nothing to do.
break;
}
// Free all the children below this node recursively.
PointerArrayUninitialize(&node->children, (PointerArrayElementFree_t)ComponentNodeFree);
free(node);
}
typedef struct ComponentTree_tag
{
ComponentNode_t root;
} ComponentTree_t;
void ComponentTreeInitialize(ComponentTree_t *tree)
{
tree->root.content.type = EMPTY;
PointerArrayInitialize(&tree->root.children);
tree->root.parent = NULL;
}
ComponentNode_t *ComponentTreeRoot(ComponentTree_t *tree)
{
return &tree->root;
}
void ComponentTreeClear(ComponentTree_t *tree)
{
if (NULL != tree)
{
PointerArrayClear(&tree->root.children,(PointerArrayElementFree_t)ComponentNodeFree);
}
}
void ComponentTreeUninitialize(ComponentTree_t *tree)
{
if (NULL != tree)
{
PointerArrayUninitialize(&tree->root.children, (PointerArrayElementFree_t)ComponentNodeFree);
}
}
ComponentNode_t *ComponentTreeCreateEmptyNode()
{
ComponentNode_t *node = (ComponentNode_t*)malloc(sizeof(ComponentNode_t));
if (NULL != node)
{
node->content.type = EMPTY;
node->parent = NULL;
PointerArrayInitialize(&node->children);
}
return node;
}
ComponentNode_t *ComponentTreeCreateNameNode(const char *name)
{
ComponentNode_t *node = (ComponentNode_t*)malloc(sizeof(ComponentNode_t));
if (NULL != node)
{
node->content.type = NAME;
node->content.name = _strdup(name);
assert(NULL != node->content.name);
if (NULL == node->content.name)
{
free(node);
return NULL;
}
node->parent = NULL;
PointerArrayInitialize(&node->children);
}
return node;
}
ComponentNode_t *ComponentTreeCreateNumberNode(float number)
{
ComponentNode_t *node = (ComponentNode_t*)malloc(sizeof(ComponentNode_t));
if (NULL != node)
{
node->content.type = NUMBER;
node->content.number = number;
node->parent = NULL;
PointerArrayInitialize(&node->children);
}
return node;
}
int ComponentTreeJoin(ComponentTree_t *tree, ComponentNode_t *where, ComponentNode_t *child)
{
if (NULL == child) return 0;
if (NULL != child->parent) return 0; // is already in a tree.
if (NULL == tree) return 0;
if (NULL == where)
{
where = &tree->root;
}
child->parent = where;
return PointerArrayPushBack(&where->children, child);
}
ComponentTree_t *tree
乍一看可能有点滑稽。参数存在于 ComponentTreeJoin()
.而是 CompoenentTreeCreateXXXNode()
函数也应该有它。为什么?因为很可能以后有人想要拥有像 ComponentTree_t *ComponentTreeFromNode(ComponentNode_t *node)
这样的功能。或 ComponentNode_t *ComponentTreeGetRoot(ComponentNode_t *node)
.最后同样重要的是,树的手动组装可能看起来像这样(不完整,这个答案已经很长了):
int _tmain(int argc, _TCHAR* argv[])
{
ComponentTree_t bikeTree;
ComponentTreeInitialize(&bikeTree);
ComponentNode_t * nutSize = ComponentTreeCreateNumberNode(0.15f);
ComponentNode_t * nut = ComponentTreeCreateNameNode("NUT");
ComponentTreeJoin(&bikeTree, nut, nutSize);
ComponentNode_t *nutCount = ComponentTreeCreateNumberNode(7.0f);
ComponentTreeJoin(&bikeTree, nutCount, nut);
ComponentNode_t * boltSize = ComponentTreeCreateNumberNode(0.1f);
ComponentNode_t * bolt = ComponentTreeCreateNameNode("BOLT");
ComponentTreeJoin(&bikeTree, bolt, boltSize);
ComponentNode_t * boltCount = ComponentTreeCreateNumberNode(5.0f);
ComponentTreeJoin(&bikeTree, boltCount, bolt);
ComponentNode_t * axle = ComponentTreeCreateNameNode("AXLE");
ComponentTreeJoin(&bikeTree, axle, nutCount);
ComponentTreeJoin(&bikeTree, axle, boltCount);
ComponentNode_t *axleCount = ComponentTreeCreateNumberNode(1.0f);
ComponentTreeJoin(&bikeTree, axleCount, axle);
ComponentNode_t *gearSize = ComponentTreeCreateNumberNode(25.0f);
ComponentNode_t *gear = ComponentTreeCreateNameNode("GEAR");
ComponentTreeJoin(&bikeTree, gear, gearSize);
ComponentNode_t *gearCount = ComponentTreeCreateNumberNode(2.0f);
ComponentTreeJoin(&bikeTree, gearCount, gear);
ComponentNode_t *hub = ComponentTreeCreateNameNode("HUB");
ComponentTreeJoin(&bikeTree, hub, gearCount);
ComponentTreeJoin(&bikeTree, hub, axleCount);
ComponentNode_t *hubCount = ComponentTreeCreateNumberNode(1.0f);
ComponentTreeJoin(&bikeTree, hubCount, hub);
ComponentNode_t * spokeSize = ComponentTreeCreateNumberNode(120.0f);
ComponentNode_t * spoke = ComponentTreeCreateNameNode("SPOKE");
ComponentTreeJoin(&bikeTree, spoke, spokeSize);
ComponentNode_t * spokeCount = ComponentTreeCreateNumberNode(1.0f);
ComponentTreeJoin(&bikeTree, spokeCount, spoke);
ComponentNode_t * axle1 = ComponentTreeCreateNameNode("AXLE");
ComponentNode_t * axle1Count = ComponentTreeCreateNumberNode(2.0f);
ComponentTreeJoin(&bikeTree, axle1Count, axle1);
// ...
ComponentNode_t * wheel = ComponentTreeCreateNameNode("WHEEL");
ComponentTreeJoin(&bikeTree, wheel, axle1Count);
ComponentTreeJoin(&bikeTree, wheel, spokeCount);
ComponentTreeJoin(&bikeTree, wheel, hubCount);
ComponentNode_t * wheelCount = ComponentTreeCreateNumberNode(2.0f);
ComponentTreeJoin(&bikeTree, wheelCount, wheel);
ComponentNode_t *bike = ComponentTreeCreateNameNode("BIKE");
ComponentTreeJoin(&bikeTree, bike, wheelCount);
ComponentNode_t *bikeCount = ComponentTreeCreateNumberNode(1.0f);
ComponentTreeJoin(&bikeTree, bikeCount, bike);
// .. frame branch omitted ..
ComponentTreeJoin(&bikeTree, NULL, bikeCount);
ComponentTreeUninitialize(&bikeTree);
_CrtDumpMemoryLeaks();
return 0;
}
概括:
main()
代码显示了如何以不完全按照解析器遍历输入文本的顺序构建树。 (我有时从右到左而不是 100% 从左到右)。 关于c - 带标准输入的 N 叉树函数(makenode、insert),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30555071/